Chemistry Prelab

Question

Pure benzaldehyde (C₇H₆O) has a normal boiling point of 178.10 °C. A solution of 24.6 g of fluorescein (C₂₀H₁₂O₅) in 128 g of benzaldehyde has a boiling point of 180.58 °C. Calculate K_bp for benzaldehyde based on this experiment.

Answer 1

For boiling point elevation (colligative property), the equation is ∆T = imK

∆T = change in boiling point = 180.58 - 178.10 = 2.48º

i = van't Hoff factor = 1 for fluorescein (a non electrolyte)

m = molality = moles fluorescein / kg benzaldehyde = see calculation below⁺

K = K_bp = ?

⁺To find m (molality) we need moles fluorescein and kg of benzaldehyde

mols fluorescein = 24.6 g x 1 mol / 332 g = 0.0741 mols

kg benzaldehyde = 128 g x 1 kg / 1000 g = 0.128 kg

m = 0.0741 mols / 0.128 kg

m = 0.579 mols / kg

Solving for K_bp we have..

∆T = imK_bp

K_bp = ∆T / (i)(m)

K_bp = 2.48º / (1)(0.579m)

K_bp = 4.28º/m